求三重积分∫dv,积分区域是由z=x^2+y^2,z=1/2*(x^2+y^2),x+y=±1,x-y=±1围成

求三重积分∫dv,积分区域是由z=x^2+y^2,z=1/2*(x^2+y^2),x+y=±1,x-y=±1围成
答案说是用换元法,但还是不清楚怎么写.

原来是极坐标变换啊,投影区域是矩形,还真有些难度的.
同样用对称性
∫∫∫Ω dV
= 4∫∫∫Ω₁ dV
= 4∫(0→1) ∫(0 → 1 - x) ∫(1/2)(x² + y²) → x² + y²) dzdydx
{ x = rcosθ
{ y = rsinθ
{ z = z
x + y = 1 ==> rcosθ + rsinθ = 1 ==> r = 1/(sinθ + cosθ)
= 4∫(0 → π/2) ∫(0 → 1/(sinθ + cosθ)) ∫(r²/2 → r²) r dzdrdθ
= 4∫(0 → π/2) ∫(0 → 1/(sinθ + cosθ)) r * (r² - r²/2) drdθ
= 4∫(0 → π/2) ∫(0 → 1/(sinθ + cosθ)) r³/2 drdθ
= 2∫(0 → π/2) [r⁴/4]：(0 → 1/(sinθ + cosθ)) dθ
= (1/2)∫(0 → π/2) 1/(sinθ + cosθ)⁴ dθ
= (1/2)∫(0 → π/2) 1/[√2sin(θ + π/4)]⁴ dθ
= (1/8)∫(0 → π/2) csc⁴(θ + π/4) dθ
= (- 1/8)∫(0 → π/2) [1 + cot²(θ + π/4)] d[cot(θ + π/4)]
= (- 1/8)[cot(θ + π/4) + (1/3)cot³(θ + π/4)]：(0 → π/2)
= - [(1/8)cot(3π/4) + (1/24)cot³(3π/4)] + [(1/8)cot(π/4) + (1/24)cot³(π/4)]
= - (- 1/8 - 1/24) + (1/8 + 1/24)
= 1/3
下面作参考：
这里有两个抛物面
其中z = x² + y²在内,2z = x² + y²在外
在xy面的投影区域D是：|x| + |y| ≤ 1
所以∫∫∫ dV = Ω的体积
= 4∫∫∫ dV = 4倍在第一挂限的体积Ω1：x + y ≤ 1
= 4∫(0 → 1) ∫(0 → 1 - x) ∫((1/2)(x² + y²) → x² + y²) dzdydx
= 4∫(0 → 1) ∫(0 → 1 - x) [x² + y² - (1/2)(x² + y²)] dydx
= 4(1/2)∫(0 → 1) ∫(0 → 1 - x) (x² + y²) dydx
= 2∫(0 → 1) [x²y + y³/3]：(0 → 1 - x) dx
= 2∫(0 → 1) [x²(1 - x) + (1 - x)³/3] dx
= (2/3)∫(0 → 1) (3x² - 3x³ + 1 - 3x + 3x² - x³) dx
= (2/3)∫(0 → 1) (- 4x³ + 6x² - 3x + 1) dx
= (2/3)[- x⁴ + 2x³ - (3/2)x² + x]：(0 → 1)
= (2/3)(- 1 + 2 - 3/2 + 1)
= 1/3